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2020年12月30日 星期三

106年台中一中教甄-數學詳解

臺中市立臺中第一高級中等學校106學年度第 1 次教師甄選

壹、填充題A部分

limnan=limnnk=12n[(2+2k2n)2+1]=20[(2+x)2+1]dx=[13x2+2x2+5x]|20=43+18=623

f2=f(f1)=f(ax+b)=a(ax+b)+b=a2x+ab+bf3=f(f2)=a3x+a2b+ab+bf7=a7+a6b+a5b++ab+b=128x+381{a7=128b(a71)=381{a=2b=3a+b=5

f(x)=x3xf(x)=3x21;P(p,p3p)3p21L:y=(3p21)(xp)+p3pL(2,a)a=(3p21)(2p)+p3pg(p)=2p36p2+2+a=0;g(p)=06p212p=06p(p2)=0p=0,2;g(p)=0g(p)=0g(0)g(2)<0(2+a)(1624+2+a)<0(a6)(a+2)<02<a<6
ABy2=4x{A(a2/4,a)B(b2/4,b);¯OA¯OBOAOB=0a2b216+ab=0ab=16OAB=12¯OA¯OB=12a416+a2b416+b2=132(a4+16a2)(b4+16b2)=132(ab)4+16a2b2(a2+b2)+162(ab)2=232+a2+b2=232+(a+b)22ab=264+(a+b)2264=16
{z1=3iz2=3i{Arg(z+3i)=Arg(zz1)Arg(z3i)=Arg(zz2)z1zz2=31040=270z1zz2{Oz1zz2r=¯z1z2÷2=3|z|=¯Oz=r=3

¯ABE¯CDF使¯EF=2CDE=¯CDׯEF÷2=3¯ABCDE¯ABsinπ3=32ABCD=3×32×13=12
x4x3+x1=x3(x1)+(x1)=(x3+1)(x1)=(x+1)(x1)(x2x+1)f(x)=x17+4x33x+1=(x+1)(x1)(x2x+1)P(x)+ax3+bx2+cx+d{f(1)=3=a+b+c+df(1)=1=a+bc+d{a+c=2b+d=1(1);x3+1f(x)x3=1f(x)x2(x3)5+4x33x+1=ax3+bx2+cx+dx23x3=bx2+cx+da{b=1c=3da=3(1){a=5b=1c=3d=25x3x23x+2


¯AD=¯DE=¯EBACD=DCE=ECB=ABC÷3=1212¯CA¯CDsinα=12¯CD¯CEsinβ=12¯CE¯CBsinγ=12¯CDsinα=¯CD¯CEsinβ=3¯CEsinγ=1sinβsinαsinγ=1/(¯CD¯CE)1/¯CD1/3¯CE=3
x31=0(x1)(x2+x+1)=0x=1,ω,ω2f(x)=(1+x+x2)1000=2000k=0akxkf(1)=31000=a0+a1+a2++a2000f(ω)=0=a0+a1ω+a2ω2+a3+a4ω+a5ω2++a2000ω2f(ω2)=0=a0+a1ω2+a2ω+a3+a4ω2+a5ω++a2000ωf(0)+f(ω)+f(ω2)=3a0+a1(1+ω+ω2)+a2(1+ω2+ω)+3a3++a2000(1+ω2+ω)=3(a0+a3+a6++a1998)31000=3666k=0a3k666k=0a3k=3999log(666k=0a3k)=log3999=999×log3=999×0.4771=476.6229666k=0a3k476+1=477
{n=1f(x)=a1x,12<x1n=2f(x)=a2x2,13<x12n=3f(x)=a3x3,14<x13n=nf(x)=anxn,1n+1<x1n;f(x){f(12)=a112=a2(12)2a2=2a1f(13)=a2(13)2=a3(13)3a3=3a2f(1n)=an1(1n)n1=an(1n)nan=nan1an=nan1=n(n1)an2==n(n1)2a1=n!
nbnbn1b2b11bn90bn1,bn2,,b19bn+bn1++b1=7Hn71=6{H16=C66=1H26=C76=7H36=C86=281+7+28=361H36=C86=2820051n=36+28+1=655n=325{H46=C96=84H56=C106=2105k=1Hk6=330>325>120=4k=1Hk6a325;{1H46=842H45=563H44=354H43=205H42=101484+56+35+20+10=20514120325;a325=552000
    (1)n=13,4,5,6;n=2512;n=3918;n=41724;n=53355×6=30<334(2)n1364/6254/3665/3676/3685/3694/36103/36112/36121/363925/2161027/2161127/2161225/2161321/2161415/2161510/216166/216173/216181/2163:p(n=1)p(n=2)p(n=3)=463036160216=100243

{{bn}bn=b1rn1{an}an=a1+(n1)d;limnnk=1bk=b11r=2+1b1=(2+1)(1r),0<r<1;{{a1=a2da3=a2+d{b1=b2/rb3=b2r{b1=a21b2=a22b3=a23{b2r=(a2d)2b2=a22b2r=(a2+d)2(a2d)2(a2+d)2=a42a42=(a22d2)2=a422a22d2+d42a22d2=d42a22=d2{d=2a2d=2a2b3=b2r=a22r=(a2+d)2{a22r=(a2+2a2)2r=(2+1)2>1a22r=(a22a2)2r=(12)2<10<r<1r=(12)2=322b1=(2+1)(1r)=(2+1)(222)=2d=2a2a1=a2d=(1+2)a2a1<a2{a1<0d>0b1=a212=a21a1=2=(1+2)a2a2=22+1=2+2d=a2a1=2+2(2)=2+22{a1=2d=2+22
0x101x0xn(1x)0xnxn1(1)1n+1=nxnxn+(1x)n+1n+1(xn)n(1x)1(n+1)n+1(xn)n(1x)nn(n+1)n+1xn(1x)=xnxn+1xnxn+1nn(n+1)n+1(2)(1)(2)0xnxn+1nn(n+1)n+1

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